anonymous
  • anonymous
1). \[\lim_{(x,y) \rightarrow (0,0)} \frac{ x ^{2}y ^{2} }{ x ^{4}+y ^{4} }\] 2). \[\lim_{(x,y) \rightarrow (0,0)} \frac{ x ^{2}y ^{2} }{ 2x ^{4}+y ^{4} }\] In the first limit I used that \[f(x,y)=f(x,0) \rightarrow 0\] and \[f(x,y)=f(0,y)\rightarrow0\] But in the second one that becomes totally wrong, since the x-axis goes to zero and y-axis to 1/3. Why should one change from f(o,y)-> 0 to x=y in the second one, how do I recognize when I should set x=y ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
experimentX
  • experimentX
*
anonymous
  • anonymous
?
experimentX
  • experimentX
just a bookmark!!

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More answers

shubhamsrg
  • shubhamsrg
Isn't answer to 1st one 1/2 ?
anonymous
  • anonymous
Nope it's 0.
experimentX
  • experimentX
it's a double limit, you have to be a bit careful with it ... since there are multiple ways to come at (0,0), I think there is a theorem .... but currently I am busty posting my own question.
shubhamsrg
  • shubhamsrg
http://www.wolframalpha.com/input/?i=limit+%28x%5E2+y%5E2+%2F%28x%5E4+%2By%5E4%29%29+as+x-%3E0+and+y-%3E0 What does this mean ?
experimentX
  • experimentX
says that limit does not exist, try y=x or y=x^2, if you don't get the same limit then limit does not exist.
anonymous
  • anonymous
According to the answerkey to Adams Calculus the answer for the first one is 0 and the second one DNE
Zarkon
  • Zarkon
\[\lim_{(x,y) \rightarrow (0,0)} \frac{ x ^{2}y ^{2} }{ x ^{4}+y ^{4} }\] let \(y=mx\) then \[\lim_{x \rightarrow 0} \frac{ x ^{2}(mx) ^{2} }{ x ^{4}+(mx) ^{4} }\] \[\lim_{x \rightarrow 0} \frac{ m^2x ^{4} }{ x ^{4}(1+m ^{4}) }\] \[\lim_{x \rightarrow 0} \frac{ m^2 }{ 1+m ^{4} }=\frac{ m^2 }{ 1+m ^{4} }\] pick any real number \(m\)

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